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(C) The equation of the sides is given by $(x^2+7xy+2y^2)(y-1)=0$. This implies the sides are $x^2+7xy+2y^2=0$ and $y-1=0$. However,$x^2+7xy+2y^2=0$ r

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$(-\frac{7}{3}, \frac{2}{3})$

Vedclass · Answer

(C) The equation of the sides is given by $(x^2+7xy+2y^2)(y-1)=0$. This implies the sides are $x^2+7xy+2y^2=0$ and $y-1=0$. However,$x^2+7xy+2y^2=0$ represents a pair of lines passing through the origin $(0,0)$. Let the lines be $L_1: y-1=0$,$L_2: y-m_1x=0$,and $L_3: y-m_2x=0$. For the lines $x^2+7xy+2y^2=0$,the sum of slopes $m_1+m_2 = -7/2$ and the product $m_1m_2 = 1/2$. The vertices are the intersection points: $V_1 = L_1 \cap L_2 \implies y=1, x=1/m_1 \implies V_1 = (1/m_1, 1)$. $V_2 = L_1 \cap L_3 \implies y=1, x=1/m_2 \implies V_2 = (1/m_2, 1)$. $V_3 = L_2 \cap L_3 \implies (0,0)$. The centroid $(G_x, G_y)$ is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$. $G_x = \frac{1/m_1 + 1/m_2 + 0}{3} = \frac{m_1+m_2}{3m_1m_2} = \frac{-7/2}{3(1/2)} = -7/3$. $G_y = \frac{1+1+0}{3} = 2/3$. Thus,the centroid is $(-\frac{7}{3}, \frac{2}{3})$.

The three sides of a triangle are given by the equation $(x^2+7xy+2y^2)(y-1)=0$. Find the centroid of the triangle.

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